Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int A[105][105], n, m;
void rotate() {
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < m; j++) {
      swap(A[i][j], A[j][i]);
    }
}
bool check() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (A[i][j]) return true;
  return false;
}
int main() {
  int R[105], t;
  bool done = true, swapped = false;
  pair<int, int> C[105];
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> A[i][j];
  if (n > m) {
    swapped = true;
    swap(n, m);
    rotate();
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      t = A[i][j];
      if (t) done = false;
      if (j == 0)
        R[i] = t;
      else
        R[i] = min(R[i], t);
      if (i == 0)
        C[j] = pair<int, int>(t, t);
      else
        C[j] = pair<int, int>(max(C[j].first, t), min(C[j].second, t));
    }
  int mxr = 0;
  bool row = false;
  vector<string> ans;
  if (!done)
    for (int i = 0; i < n; i++)
      if (R[i]) {
        row = true;
        string tmp = swapped ? "col " : "row ";
        tmp += to_string(i + 1);
        for (int k = 0; k < R[i]; k++) {
          for (int j = 0; j < m; j++) A[i][j]--;
          ans.push_back(tmp);
        }
        mxr = max(mxr, R[i]);
      }
  if (!done)
    for (int j = 0; j < m; j++) {
      int c = row ? C[j].first - mxr : C[j].second;
      if (c) {
        string tmp = swapped ? "row " : "col ";
        tmp += to_string(j + 1);
        for (int k = 0; k < c; k++) {
          for (int i = 0; i < n; i++) A[i][j]--;
          ans.push_back(tmp);
        }
      }
    }
  if (check() || (!ans.size() && !done))
    cout << -1 << endl;
  else if (done)
    cout << 0 << endl;
  else {
    cout << ans.size() << endl;
    for (string s : ans) cout << s << endl;
  }
}